Proper Factors

Show that $3^2\times 5^3$ has exactly $10$ proper factors. Determine how many other integers of the form $3^m\times5^n$ (where $m$ and $n$ are integers) have exactly $10$ proper factors. 

Dylan from Brooke Weston and Ishaan from Simon Balle All-Through School, both in the UK, used factorisations to count the number of proper factors of $3^2\times5^3.$ This is Dylan's work:

Prem from Cranford Community College in England used a table to count the proper factors, and to find a formula for the total number of proper factors:

To find the amount of ”co-ordinates” on our graph we can simply do our length $\times$ width but we have to remember to subtract $2$ since a proper factor cannot be $1$ or $N.$

This makes the second part rather easier to visualise as we extend both axis of our graph to an arbitrary pair of values of $m, n.$ The total possible co-ordinates is still given by the same formula - our length would be $m + 1$ and our width would be $n + 1$ giving the total possible amount of pairs as $(m + 1)(n + 1)$ but as before, we have to subtract the top left and bottom right pairs as they would multiply to give us $1$ and $N$ which does not satisfy the conditions of a proper factor. $\Rightarrow$ the amount of proper factors $z = (m + 1)(n + 1) − 2.$

Nishad from Thomas Estley Community College in the England, Xuan Tung from HUS High School for Gifted Students in Vietnam, Natal from Canada, Dylan and Ishaan found the same formula without using a table. This is Xuan Tung's work:

Joshua from Bohunt Sixth Form in the UK, Ong from Kelvin Grove State College Brisbane in Australia, Nishad, Xuan Tung, Natal, Prem, Dylan and Ishaan used this formula - or these ideas - to count how many other numbers of the form $3^m\times5^n$ have $10$ proper factors. This is Ong's work:

There are two more possibilities that Ong missed but Nishad found:

So $3^0\times5^{11}$ and $3^{11}\times5^0$ also have $10$ proper factors.

Ziwei (Gilbert) from Stowe School in the UK and Dibyadeep from Greenhill School in the USA counted up the factors in a different way. This is Ziwei (Gilbert)'s work (click on the images to open larger versions):

Let $N$ be the smallest positive integer that has exactly $426$ proper factors. Determine $N$, giving your answer in terms of its prime factors. 

Joshua, Ishaan and Ong used the formula to find $N.$ They all assumed that $N$ would have three distinct prime factors. This is Ishaan's work:

Nishad, Dibyadeep, Xuan Tung, Dylan and Natal used a similar method, but they did not assume anything about the number of distinct prime factors of $N.$ This is Dibyadeep's work:

Dylan's method was shorter:

Natal explained why $N$ should have three distinct prime factors. This is Natal's work (click on the images to open larger versions):